JEE Main 2015 (Online) 10th AprilQuestion 1
If the points $(1,1,\lambda)$ and $(-3,0,1)$ are equidistant from the plane $3x+4y-12z+13=0$, then $\lambda$ satisfies the equation:
3D Geometry
Solution
$\frac{|3+4-12\lambda+13|}{13}=\frac{|-9-12+13|}{13}$ gives $|20-12\lambda|=8$, $\lambda=1$ or $\frac73$. Sum $\frac{10}3$, product $\frac73$: $3\lambda^2-10\lambda+7=0$.